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ESATIMATION:
We represent two formulas for dividing
the mean and the standard deviation of the
binomial distribution:
μ=np
s=√npq
where
 n= number of trails
 p= probability of success
 q=1-p=probability of failures
Theoretically, the binomial distribution is the
correct distribution is the distribution to use in
constructing confidence intervals to estimate a
population proportion.
As the sample size increases, the binomial
can be approximate by an appropriate normal
distribution, which we can use to approximate
the sampling distribution. Statisticians
recommend that in estimation, n be the large
enough for both np and nq to be at least 5 when
n you can use the normal distribution as a
substitute for the binomial.
Let’s express the proportion of
successes in a sample by p (p bar). Then modify
equation 5-2,so that we can use it to drive the
mean of the sampling distribution of the
proportion of successes. In words, μ=np shows
that the mean of the binomial distribution is
equal to the product of number of trails, n, and
the probability of successes; that is, np equals
the mean number of successes. To change this
number of successes to the proportion of
successes, we divide np by n and get p alone.
The mean in the left _hand side of the equation
becomes μp, or the mean of sampling
distribution of the proportion of successes.
MEAN OF SAMPLING DISTRIBUTION
OFR RHE PROPORTION
μp=p…………………….[7-3]
Similarly, we can modify the formula for the
standard deviation of the binomial distribution, √npq,
which measures the standard deviation in the number
of successes ,we divide √npq by n and get √pq/n. In
statistical terms, the standard deviation for the
proportion of successes in a sample is symbolized
and is called the standards error of the proportion.
STANDARD ERROR OF THE
PROPORTION
Standard error of the proportion
……………….. [7-4]
We can illustrate how to use these formulas if we
estimate for a very organization what proportion of
the employees prefer to provide their own retirement
benefits in lieu of a company _sponsored plan. First,
we conduct a simple random sample of 75 employees
and find that 0.4 of them are interested in providing
their own plans. Our results are
n=75…..sample size
p=0.4…..sample proportion
Next, management requests that we can use
this sample to find an interval about which
they can be 99% confident that it contains
the true population proportion.
But what are p and q for the population?
We can estimate the population parameters
by substituting the corresponding sample
statistics p and q (p bar and q bar) in the
formula for the standard error of the
proportion.*Doing this we get
Now we can provide the estimate
management needs by using the same
procedure we have used previously. A 99%
confidence level would include 49.5% of the
area on either side of the mean in sampling
distribution. The body of appendix Table 1
tells us that 0.495 of the area under the
normal curve located b/w the mean and a
point 2.58 standard errors from the mean.
Thus, 99% of the area is contained b/w plus
and minus 2.58 standard error from the
mean. Our confidence limits then becomes
P+2.58 Ồp
=0.4+2.58(0.057)
=0.4+0.147
=0.547……….upper confidence limit
p-2.58 Ồp=0.42.58(0.057)
=0.4 –
0.147
=0.253…………lower confidence limit
thus, we estimate from our sample
of 75 employees that with 99% confidence
we believe that the proportion of the total
population of employees who wish to their
own retirement plans lies b/w 0.253 and
0.547.